# Prove That No Polynomial Sends $\sin(\theta)$ to $\cos(\theta)$ For All $\theta$ In Some Non-Empty Interval

Note: In the context of this question, when I say "polynomial", I mean a single variable polynomial with coefficients in $$\mathbb{R}$$ and one that has $$\mathbb{R}$$ as the domain and codomain.

Question: Prove that there exists no polynomial $$p$$ such that $$p(\sin(\theta))=\cos(\theta)\forall\theta\in I,$$ where $$I$$ is some non-empty interval of $$\mathbb{R}.$$

My Attempt: For the sake of contradiction, let such a polynomial exist. Then, define $$g(x):=p(x)^2+x^2-1.$$ Since $$p$$ is a polynomial, it follows that $$g$$ is too. Now, for all $$a\in I, g(\sin(a))=0.$$ Since the interval is non-empty, this occurs infinitely many times. By the fundamental theorem of algebra, any polynomial that is zero infinitely many times must be zero identically. Hence, $$g(x)=0\forall x\in\mathbb{R}.$$ This means that $$p(x)=±\sqrt{1-x^2}\forall x\in\mathbb{R}.$$ Now, $$\sqrt{2}\in\mathbb{R},$$ but $$p(\sqrt{2})=±i\notin\mathbb{R}.$$ This is in contradiction with the fact that the codomain of $$p$$ is $$\mathbb{R}.$$ Hence, the initial assumption about the existence of such a polynomial we made must be false. This shows that no such polynomial exists.

Is my argument right? I fear I've made some conceptual error that invalidates my entire proof. Can this statement be proven in a different way?

• An easy way out is $p(\sin(0)) = \cos(0) = 1 \neq -1 = \cos(\pi) = p(\sin(\pi))$, but I suppose this is just an oversight of the question statement. Commented Feb 8, 2023 at 0:13
• @L.F., good catch. I need to edit the question to clarify this. The actual question says "for all $\theta$ in some non-empty interval". Thankfully this doesn't change my proof by much. Commented Feb 8, 2023 at 0:16
• @Mike, sorry for my negligence. I didn't mean to annoy, and I'll edit it soon. Commented Feb 8, 2023 at 0:22
• @Mike, all good now, man? Tbh, I don't see why you were so annoyed. Mistakes happen, you know? Commented Feb 8, 2023 at 0:30
• Mistakes cannot happen if you are designing a bridge :) It is on the person writing to be clear on their question from the beginning!
– Mike
Commented Feb 8, 2023 at 0:32

If $$x= \sin t$$ and $$P(x)= \cos t$$ then by the Chain Rule $$P'(x) \cos t= -\sin t$$ so $$P'(x) P(x)=-x$$ for some interval of values for $$x$$. But the degrees of the polynomials on left and right cannot match unless that degree is 1. What happens then?

• Very nice! I suspected that the argument had something to do with degrees but this was very slick. +1
– Mike
Commented Feb 8, 2023 at 1:00
• This implies that $p$ is a linear polynomial, which is easy to disprove. Thanks. Commented Feb 8, 2023 at 1:08

By the fundamental theorem of algebra, any polynomial that is zero infinitely many times must be zero identically. Hence, $$g(x)=0\forall x\in\mathbb{R}.$$ This means that $$p(x)=±\sqrt{1-x^2}\forall x\in\mathbb{R}.$$ Now, $$\sqrt{2}\in\mathbb{R},$$ but $$p(\sqrt{2})=±i\notin\mathbb{R}.$$ This is in contradiction with the fact that the codomain of $$p$$ is $$\mathbb{R}.$$

This part of the proof is a little dodgy, and the issue is domains (rather than codomains). You've proven that $$g(x) = 0$$ for all $$x \in \mathbb{R}$$ but then you identify $$p(x)$$ with a function that, regarded as a real function, has domain $$[-1, 1]$$ (and then you evaluate it outside of its domain), which is... I won't say incorrect, but it's a little awkward and potentially confusing.

I think a cleaner way to argue from here is just to prove that there is no real polynomial $$p(x)$$ satisfying $$p(x)^2 = 1 - x^2$$. There are many ways to do this:

1. $$p(x)$$ must be a linear polynomial, but then the leading coefficient of $$p(x)^2$$ must be positive, so can't be $$-1$$.
2. $$p(\sqrt{2})^2 = 1 - 2 = -1$$ (this is essentially your proof but we avoid the complication mentioned above with domains by keeping the square) which is a contradiction.
3. $$1 - x^2 = (1 - x)(1 + x)$$ is not a square because it has two distinct roots (this argument proves that $$p(x)$$ can't be a complex polynomial either).
• Thanks, this clears a lot of the confusion I had with my argument. Personally, I prefer method $2$ because it's so similar in spirit to my original argument. Commented Feb 8, 2023 at 2:40

Assume WLOG that $$I$$ is such that $$\sin(\theta)$$ is either positive or negative for all $$\theta \in I$$, and that $$\cos(\theta)$$ is either positive or negative for all $$\theta \in I$$. So then this partitions into 4 cases:

1. $$\sin(\theta)$$ positive for all $$\theta \in I$$ and $$\cos(\theta)$$ positive for all $$\theta \in I$$.

2. $$\sin(\theta)$$ positive for all $$\theta \in I$$ and $$\cos(\theta)$$ negative for all $$\theta \in I$$.

3. $$\sin(\theta)$$ negative for all $$\theta \in I$$ and $$\cos(\theta)$$ positive for all $$\theta \in I$$.

4. $$\sin(\theta)$$ negative for all $$\theta \in I$$ and $$\cos(\theta)$$ negative for all $$\theta \in I$$.

Let us now consider Case 1: $$\sin(\theta)$$ positive for all $$\theta \in I$$ and $$\cos(\theta)$$ positive for all $$\theta \in I$$. The remaining 3 cases can be handled similarly.

Then for Case 1, let $$J$$ be the set $$\{x:$$ there is a $$\theta \in I$$ such that $$\sin(\theta)=x\}$$. Then $$J$$ is a nonempty interval on the reals. Then $$\sqrt{1-p^2(x)} = x \ \forall x \in J$$, which gives, squaring both sides $$1-p^2(x) = x^2 \ \forall x \in J$$. Which gives [as $$p^2(x)$$ is a polynomial and $$J$$ is infinite cardinality] $$p^2(x) = x^2-1$$ $$\ \forall x \in \mathbb{R}$$, which gives deg$$(p^2)=$$ $$2$$deg$$(p)$$ $$=$$ deg$$(x^2-1)$$ $$=2$$, which gives deg$$(p)=1$$. So then $$p$$ must be a linear polynomial, which is absurd.

So you were close in your OP.

• Thanks for your input. Is my original proof right, though? Commented Feb 8, 2023 at 1:07
• You were close until the ending. Where there may be problems with your argument is that you concluded that if $p(x)=q(x)$ on an interval; $q$ in your proof $q(x)=\sqrt{1-x^2}$ *not * a polynomial, then $p$ and $q$ must be equal everywhere. [This is true if $q$ is a polynomial, as you noted already.] I'm not sure how the graders will take what you wrote though.
– Mike
Commented Feb 8, 2023 at 1:21
• You could have just concluded that $1-p^2(x) =x^2$, and then [as $x^2$ is a polynomial] that $1-p^2(x)=x^2$ everywhere, and then conclude that $p^2$ has degree $2$ then so $p$ must have degree $1$ and thus be linear.
– Mike
Commented Feb 8, 2023 at 1:25
• yes I now see the potential issues with my proof. I'll keep this sort of degree argument in mind, thanks. Commented Feb 8, 2023 at 2:43